University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 72



Work Step by Step

We know that $\sin x$ is a periodic function with period $\dfrac{2\pi}{k}$. Now, $A=\int_0^{\pi/k} \sin kx dx$ This implies that $[(\dfrac{-1}{k}) \cos kx]_0^{\pi/k}=\dfrac{-1}{k}\cos k(1/k)-(\dfrac{-1}{k}\cos 0)$ Thus, $A=\dfrac{2}{k}$
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